Question #d8c45

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Answer 1 · 2016-07-27T15:44:36

Given

$costheta+sintheta=sqrt2costheta$

Following the formula

$(a+b)^2+(a-b)^2=2(a^2+b^2)$

we can wite

$(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)$

Putting

$costheta+sintheta=sqrt2costheta" "$ we get

$(sqrt2costheta)^2+(costheta-sintheta)^2=2*1$

$=>2cos^2theta+(costheta-sintheta)^2=2*1$

$=>(costheta-sintheta)^2=2-2cos^2theta$

$=>(costheta-sintheta)^2=2(1-cos^2theta)$

$=>(costheta-sintheta)^2=2sin^2theta$

$=>costheta-sintheta=sqrt2sintheta$

Proved

In other way when
$costheta-sintheta=sqrt2sintheta" "$ we get

$(costheta+sintheta)^2+(sqrt2sintheta)^2=2*1$

$=>(costheta+sintheta)^2 + 2sin^2theta =2$

$=>(costheta+sintheta)^2=2-2sin^2theta$

$=>(costheta+sintheta)^2=2(1-sin^2theta)$

$=>(costheta+sintheta)^2=2cos^2theta$

$=>costheta+sintheta=sqrt2costheta$

Proved